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2 Learning Objectives for Section 4.3 Gauss-Jordan EliminationThe student will be able to convert a matrix to reduced row echelon form.The student will be able to solve systems by Gauss-Jordan elimination.The student will be able to solve applications using Gauss-Jordan elimination.Barnett/Ziegler/Byleen Finite Mathematics 12e

3 Gauss-Jordan EliminationAny linear system must have exactly one solution, no solution, or an infinite number of solutions.Previously we considered the 2  2 case, in which the term consistent is used to describe a system with a unique solution, inconsistent is used to describe a system with no solution, and dependent is used for a system with an infinite number of solutions. In this section we will consider larger systems with more variables and more equations, but the same three terms are used to describe them.Barnett/Ziegler/Byleen Finite Mathematics 12e

4 Carl Friedrich GaussAt the age of seven, Carl Friedrich Gauss started elementary school, and his potential was noticed almost immediately. His teacher, Büttner, and his assistant, Martin Bartels, were amazed when Gauss summed the integers from 1 to 100 instantly by spotting that the sum was 50 pairs of numbers, each pair summing to 101.Barnett/Ziegler/Byleen Finite Mathematics 12e

5 Matrix Representations of Consistent, Inconsistent and Dependent SystemsThe following matrix representations of three linear equations in three unknowns illustrate the three different cases:From this matrix representation, you can determine that x = 3 y = 4 z = 5Case I: consistentBarnett/Ziegler/Byleen Finite Mathematics 12e

6 Matrix Representations (continued)Case 2: inconsistentFrom the second row of the matrix, we find that0x + 0y +0z =6or0 = 6,an impossible equation. From this, we conclude that there are no solutions to the linear system.Barnett/Ziegler/Byleen Finite Mathematics 12e

7 Matrix Representations (continued)Case 3: dependentWhen there are fewer non-zero rows of a system than there are variables, there will be infinitely many solutions, and therefore the system is dependent.Barnett/Ziegler/Byleen Finite Mathematics 12e

8 Reduced Row Echelon FormA matrix is said to be in reduced row echelon form or, more simply, in reduced form, ifEach row consisting entirely of zeros is below any row having at least one non-zero element.The leftmost nonzero element in each row is 1.All other elements in the column containing the leftmost 1 of a given row are zeros.The leftmost 1 in any row is to the right of the leftmost 1 in the row above.Barnett/Ziegler/Byleen Finite Mathematics 12e

12 Example (continued)We already have a 1 in the diagonal position of first column. Now we want 0’s below the 1. The first 0 can be obtained by multiplying row 1 by –2 and adding the results to row 2:Row 1 is unchanged(–2) times Row 1 is added to Row 2Row 3 is unchangedBarnett/Ziegler/Byleen Finite Mathematics 12e

13 Example (continued)The second 0 can be obtained by adding row 1 to row 3:Row 1 is unchangedRow 2 is unchangedRow 1 is added to Row 3Barnett/Ziegler/Byleen Finite Mathematics 12e

14 Example (continued)Moving to the second column, we want a 1 in the diagonal position (where there was a –3). We get this by dividing every element in row 2 by -3:Row 1 is unchangedRow 2 is divided by –3Row 3 is unchangedBarnett/Ziegler/Byleen Finite Mathematics 12e

15 Example (continued)To obtain a 0 below the 1 , we multiply row 2 by –3 and add it to the third row:Row 1 is unchangedRow 2 is unchanged(–3) times row 2 is added to row 3Barnett/Ziegler/Byleen Finite Mathematics 12e

16 Example (continued)To obtain a 1 in the third position of the third row, we divide that row by 4. Rows 1 and 2 do not change.Barnett/Ziegler/Byleen Finite Mathematics 12e

17 Example (continued)We can now work upwards to get zeros in the third column, above the 1 in the third row.Add R3 to R2 and replace R2 with that sumAdd R3 to R1 and replace R1 with the sum.Row 3 will not be changed.All that remains to obtain reduced row echelon form is to eliminate the 1 in the first row, 2nd position.Barnett/Ziegler/Byleen Finite Mathematics 12e

18 Example (continued)To get a zero in the first row and second position, we multiply row 2 by –1 and add the result to row 1 and replace row 1 by that result. Rows 2 and 3 remain unaffected.Barnett/Ziegler/Byleen Finite Mathematics 12e

19 Final ResultWe can now “read” our solution from this last matrix. We havex = 1, y = –1 z = 2.Written as an ordered triple, we have (1, –1, 2). This is a consistent system with a unique solution.Barnett/Ziegler/Byleen Finite Mathematics 12e

24 Final ResultTo get a zero in the third row, second entry we multiply row 2 by –1 and add the result to R3 and replace R3 by that sum: Notice this operations “wipes out” row 3 so row 3 consists entirely of zeros.Any time you have fewer non-zero rows than variables you will have a dependent system.Barnett/Ziegler/Byleen Finite Mathematics 12e

25 Representation of a Solution of a Dependent SystemNext we can express the variable x in terms of t as follows: From the first row of the matrix, we have x – y -2z = 2. If z = t and y = 10t – 1, we have x – (10t – 1) – 2t = 2 or x = 12t + 1.Our general solution can now be expressed in terms of t: (12t + 1,10t – 1, t), where t is an arbitrary real number.We can interpret the second row of this matrix as –y + 10z = 1, or z – 1 = ySo, if we let z = t (arbitrary real number,) then in terms of t, y = 10t – 1.Barnett/Ziegler/Byleen Finite Mathematics 12e

26 Procedure for Gauss-Jordan EliminationStep 1. Choose the leftmost nonzero column and use appropriate row operations to get a 1 at the top.Step 2. Use multiples of the row containing the 1 from step 1 to get zeros in all remaining places in the column containing this 1.Step 3. Repeat step 1 with the submatrix formed by (mentally) deleting the row used in step and all rows above this row.Step 4. Repeat step 2 with the entire matrix, including the rows deleted mentally. Continue this process until the entire matrix is in reduced form.Note: If at any point in this process we obtain a row with all zeros to the left of the vertical line and a nonzero number to the right, we can stop because we will have a contradiction.Barnett/Ziegler/Byleen Finite Mathematics 12e

27 ApplicationsSystems of linear equations provide an excellent opportunity to discuss mathematical modeling. The process of using mathematics to solve real-world problems can be broken down into three steps:Step 1. Construct a mathematical model whose solution will provide information about the real-world problem.Real-world problem3. Interpret1. ConstructMathematical SolutionMathematical Model2. SolveBarnett/Ziegler/Byleen Finite Mathematics 12e

28 Applications (continued)Step 2. Solve the mathematical model.Step 3. Interpret the solution to the mathematical model in terms of the original real-world problem.Example: Purchasing. A company that rents small moving trucks wants to purchase 25 trucks with a combined capacity of 28,000 cubic feet. Three different types of trucks are available: a 10-foot truck with a capacity of 350 cubic feet, a 14-foot truck with a capacity of 700 cubic feet, and a 24-foot truck with a capacity of 1,400 cubic feet. How many of each type of truck should the company purchase?Barnett/Ziegler/Byleen Finite Mathematics 12e

29 SolutionStep 1. The question in this example indicates that the relevant variables are the number of each type of truck.x = number of 10-foot trucksy = number of 14-foot trucksz = number of 24-foot trucksWe form the mathematical model:x + y + z = (Total number of trucks)350x + 700y + 1,400z = 28,000 (Total capacity)Barnett/Ziegler/Byleen Finite Mathematics 12e

31 Solution (continued) Let z = t. Then for t any real numberx = 2t – 30 y = –3t z = tis a solution to our mathematical model.Step 3. We must interpret this solution in terms of the original problem. Since the variables x, y, and z represent numbers of trucks, they must be nonnegative. And since we can’t purchase a fractional number of trucks, each must be a nonnegative whole number.Barnett/Ziegler/Byleen Finite Mathematics 12e

32 Solution (continued)Since t = z, it follows that t must also be a nonnegative whole number. The first and second equations in the model place additional restrictions on the values t can assume:x = 2t – 30 > 0 implies that t > 15 y = –3t + 55 > 0 implies that t < 55/3Thus the only possible values of t that will produce meaningful solutions to the original problem are 15, 16, 17, and 18. A table is a convenient way to display these solutions.10-ft truck14-ft. truck24-ft trucktxyz151016271741861Barnett/Ziegler/Byleen Finite Mathematics 12e

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